A) \[\frac{2\pi }{2}\sec \]
B) \[\pi \sec \]
C) \[\frac{3\pi }{2}\sec \]
D) \[2\pi \sec \]
Correct Answer: B
Solution :
For a body executing SHM, velocity,\[v=\sqrt{{{\omega }^{2}}({{a}^{2}}-{{y}^{2}})}\]we have \[{{10}^{2}}={{\omega }^{2}}({{a}^{2}}-{{4}^{2}})\]and \[{{8}^{2}}={{\omega }^{2}}({{a}^{2}}-{{5}^{2}})\] So, 1\[{{10}^{2}}-{{8}^{2}}={{\omega }^{2}}({{5}^{2}}-{{4}^{2}})={{(3\omega )}^{2}}\]or \[6=3\omega \] or \[\omega =2\] \[\because \] Time, \[t=2\pi /\omega \] \[\therefore \] \[t=2\pi /2=\pi \sec \]
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